Let $M$ be closed subspace of Hilbert space $H$ and let $f$ be bounded linear functional on $M$. Prove that there is unique expansion $F$ from $f$ on whole $H$ which satisfies $||F||=||f||$. Prove that $F=0$ on $M^\bot$.
From Hahn-Banach theorem we know that there is such expansion, but it doesn't guarantee uniqueness. Since $M$ is closed, we can utilize theorem of orthogonal projection. I guess we need to suppose that there are two expansions with given property, but what next?
As $M$ is a closed subspace of $H$, it is itself a Hilbert space. By Riesz representation theorem there is $x_0 \in M$ such that $$f(x) = (x_0, x)_H, \quad \forall x \in M.$$ Now define $F$ by $$F(x) = (x_0, x)_H, \quad \forall x \in H.$$ The uniqueness of the extension comes from the uniqueness of $x_0$ (and it is clearly zero on $M^\perp$ as $x_0 \in M$).